Circuit Analysis 2: AC circuits: cram pack

Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Complex numbers

Rectangular form
z = a + jb, j² = −1, z̄ = a − jb
Modulus and argument
|z| = √(a² + b²), tan θ = b/a (check the quadrant)
Polar and exponential form
z = r∠θ = r(cos θ + j sin θ) = r·ejθ
Division
z₁/z₂ = z₁z̄₂/|z₂|²
Product and quotient
r₁∠θ₁ · r₂∠θ₂ = r₁r₂∠(θ₁ + θ₂), r₁∠θ₁ / r₂∠θ₂ = (r₁/r₂)∠(θ₁ − θ₂)
De Moivre
(r∠θ)ⁿ = rⁿ∠nθ
n-th roots
zk = R1/n∠((φ + 360°·k)/n), k = 0 … n − 1
Quadratic with p² < 4q
z = −p/2 ± j√(q − p²/4)

Method

  1. Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
  2. Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
  3. Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
  4. For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).

AC impedance & phasors

Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z

Method

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

AC power & power factor

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

LC resonance, Q factor & bandwidth

Natural frequency
ω0 = 1/√(LC), f0 = 1/(2π√(LC))
Energy
W = ½CV0² = ½Cv² + ½Li²
Peak current
Imax = V0√(C/L)
Quality factor
Q = 2π·(energy stored)/(energy lost per cycle)
Series RLC
Q = (1/R)√(L/C), BW = R/(2πL)
Parallel RLC
Q = R√(C/L), BW = 1/(2πRC)
Bandwidth
BW = f2 − f1 = f0/Q, f0 = √(f1f2)
Half-power frequencies
f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q))
Tank with coil resistance r
RD = L/(C·r) = Q²·r
Decay of the amplitude
e−Rt/(2L)

Method

  1. Find the resonance frequency first: ω₀ = 1/√(LC) and f₀ = ω₀/(2π).
  2. Decide whether the circuit is series or parallel: series Q = (1/R)√(L/C), parallel Q = R√(C/L).
  3. Use BW = f₀/Q for the bandwidth, or the shortcuts R/(2πL) (series) and 1/(2πRC) (parallel).
  4. For energy questions, use that the total energy ½CV₀² stays constant in an ideal LC circuit and swings between C and L every quarter period.

Three-phase power

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

RC & RL transients

Time constant
τ = RC | τ = L/R
Response
x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
Capacitor at DC
open circuit; vC cannot jump
Inductor at DC
short circuit; iL cannot jump
Settled
t ≈ 5τ

Method

  1. Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
  2. Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
  3. Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
  4. Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.

Thévenin & Norton

Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)

Method

  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.

2Practice problems

1.Complex numbersMedium
ReIm11−1−122−2−233−3−3θz

Find the argument θ of z = −3 − j3 in degrees, as the principal value between −180° and 180°.

z = −3 − j3
θ = °
2.Complex numbersHard
ReIm11−1−122−2−2z₀z₁

The roots of zⁿ = w with n = 2 and w = 4∠90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the imaginary part of z₀.

w = 4∠90°n = 2k = 0
Im(z₀) =
3.AC impedance & phasorsMedium
120 V500 HzR = 15 ΩC = 47 µFI

A resistor and a capacitor are connected in parallel across a 120 V (rms), 500 Hz supply. Find the magnitude of the impedance |Z|.

V = 120 V (rms)f = 500 HzR = 15 ΩC = 47 µF
|Z| = Ω
4.AC impedance & phasorsHard
2 Vf0​R = 50 ΩL = 5 mHC = 0.1 µF

A series RLC circuit is driven by a 2 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 2 V (rms)R = 50 ΩL = 5 mHC = 0.1 µF
VL = V
5.AC power & power factorMedium
400 V116 kW, 0.65213 kW, 1I

Two loads are connected in parallel to a 400 V single-phase supply: a 16 kW load at a power factor of 0.65 lagging and a 13 kW heater (unity power factor). Find the overall power factor cos φ seen by the supply.

V = 400 VP1 = 16 kW, pf1 = 0.65 laggingP2 = 13 kW, pf2 = 1
cos φ =
6.AC power & power factorHard
240 V60 Hzload17 kW, 0.6C330 µFI

A single-phase load takes 17 kW at a power factor of 0.6 lagging from a 240 V, 60 Hz supply. A 330 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 17 kWpf1 = 0.6V = 240 Vf = 60 HzC = 330 µF
cos φ =
7.LC resonance, Q factor & bandwidthMedium
tvC​iL​T/4 = ?T/2TL = 10 µHC = 10 nF

In an ideal LC circuit with L = 10 µH and C = 10 nF, the capacitor is fully charged at t = 0. How long does it take until all the energy is in the inductor? Give t in µs.

L = 10 µHC = 10 nF
t = µs
8.LC resonance, Q factor & bandwidthHard
V100 ΩLC = ?

Design a series RLC filter with a total resistance of 100 Ω, a resonance frequency of 10 kHz and a bandwidth of 1 kHz. Find the capacitance C in nF.

R = 100 Ωf0 = 10 kHzBW = 1 kHz
C = nF
9.Three-phase powerMedium
ABCVL​ = 400 VNZ = 9 + j12 Ωper phase

A balanced Y-connected load with an impedance of 9 + j12 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 400 V. Find the total real power P absorbed by the load in kW.

VL = 400 VZ = 9 + j12 Ω per phase
P = kW
10.Three-phase powerHard
P = 120 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 120 kW at a power factor of 0.78 lagging from a 480 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 120 kWpf1 = 0.78pf2 = 0.9VL = 480 Vf = 50 Hz
QC = kvar
11.RC & RL transientsMedium
R = 33 Ωt = 0L220 mHi(t)

The inductor carries a steady current of 3 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. Find the current i at t = 15 ms.

i(0) = 3 AR = 33 ΩL = 220 mHt = 15 ms
i = A
12.RC & RL transientsHard
+−Vs​12 Vopens at t = 0R1​ = 4.7 kΩR2​ = 1 kΩC220 µF+−vC​

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage vC at t = 500 ms.

Vs = 12 VR1 = 4.7 kΩR2 = 1 kΩC = 220 µFt = 500 ms
vC = V
13.Thévenin & NortonMedium
R1​ = 3 ΩR3​ = 4 Ω+−V1​9 VR2​12 ΩAab

Find the Thévenin resistance Rth seen from terminals a–b.

V1 = 9 VR1 = 3 ΩR2 = 12 ΩR3 = 4 Ω
Rth = Ω
14.Thévenin & NortonHard
R1​ = 25 ΩR3​ = 10 Ω+−V1​30 VR2​8 ΩAIs​3 ABRL​20 ΩIL​

A load RL = 20 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 30 VIs = 3 AR1 = 25 ΩR2 = 8 ΩR3 = 10 ΩRL = 20 Ω
IL = A
15.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−60−40−200204060ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 2

Find the phase ∠G(jω) of G(s) = 7(s + 20)/(s(s + 7)) at ω = 2 rad/s, in degrees.

G(s) = 7(s + 20)/(s(s + 7))ω = 2 rad/s
∠G(jω) = °
16.Bode plots & marginsHard
|H| (dB)11101010010010001000−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−180−135−90−450ωc​ = ?

A unity-feedback loop has the open-loop transfer function L(s) = 250/(s(s + 10)). Find the gain crossover frequency ωc, where |L(jω)| = 1.

L(s) = 250/(s(s + 10))
ωc = rad/s

3Solutions

1.Complex numbers

Answer: θ = −135 °

Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.

Quadrant

a = −3, b = −3 → III

Reference angle

α = atan(3/3) = 45°

Argument

θ = −(180° − α) = −135°

Answer

θ = −135°

In polar form z = 4.243∠−135°.

2.Complex numbers

Answer: Im(z₀) = 1.414

De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.

Modulus of the roots

41/2 = 2

Argument of the root

θ₀ = (90° + 360°·0)/2 = 45°

Rectangular form

z₀ = 2·(cos 45° + j sin 45°) = 1.414 + j1.414
Im(z₀) = 1.414

Answer

Im(z₀) = 1.414

The 2 roots lie on the circle of radius 2, 180° apart.

3.AC impedance & phasors

Answer: |Z| = 6.173 Ω

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Reactance

XC = 1/(ωC) = 1/(2π·500·47·10⁻⁶) = 6.773 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

IR = V/R = 120/15 = 8 A
IX = V/XC = 120/6.773 = 17.72 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(IR² + IX²) = √(8² + 17.72²) = 19.44 A

Impedance

|Z| = V/I = 120/19.44 = 6.173 Ω

Answer

|Z| = 6.173 Ω
4.AC impedance & phasors

Answer: VL = 8.944 V

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.005·0.1·10⁻⁶) = 44721 rad/s
I = V/R = 2/50 = 0.04 A

Voltage across the inductor

XL = ω₀L = 44721·0.005 = 223.6 Ω
VL = I·XL = 0.04·223.6 = 8.944 V

Answer

VL = 8.944 V

This is Q = 4.472 times the supply voltage.

5.AC power & power factor

Answer: cos φ = 0.8403

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 16·tan(arccos 0.65) = 18.71 kvar
Q2 = 0 = 0 kvar

Totals

P = 16 + 13 = 29 kW
Q = 18.71 + 0 = 18.71 kvar
S = √(29² + 18.71²) = 34.51 kVA

Overall power factor

cos φ = P/S = 29/34.51 = 0.8403 lagging

Answer

cos φ = 0.8403
6.AC power & power factor

Answer: cos φ = 0.7389

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = 17·tan(arccos 0.6) = 22.67 kvar
QC = V²·2πf·C = 240²·2π·60·330·10⁻⁶ = 7166 var = 7.166 kvar
Q2 = 22.67 − 7.166 = 15.5 kvar

New apparent power

S2 = √(17² + 15.5²) = 23.01 kVA

New power factor

cos φ = P/S2 = 17/23.01 = 0.7389

Answer

cos φ = 0.7389
7.LC resonance, Q factor & bandwidth

Answer: t = 0.4967 µs

LC resonance: a charged capacitor connected to an inductor drives a current that charges it the other way: the energy swings between the electric field of C (½Cv²) and the magnetic field of L (½Li²) at the natural frequency f0 = 1/(2π√(LC)). Without resistance the total energy ½CV0² stays constant, so the peak current is Imax = V0√(C/L). One period is T = 1/f0; the energy moves from C to L in a quarter of it.

Period

T = 2π√(1 × 10⁻⁵·1 × 10⁻⁸) = 1.987 µs

Quarter period

t = T/4 = 0.4967 µs

Answer

t = 0.4967 µs
8.LC resonance, Q factor & bandwidth

Answer: C = 15.92 nF

Q factor and bandwidth: in a resonant circuit Q = 2π·(energy stored)/(energy lost per cycle). For series RLC Q = ω0L/R = (1/R)√(L/C); for parallel RLC Q = R/(ω0L) = R√(C/L). The bandwidth between the half-power frequencies f1 and f2 (where the power is half and the current or voltage 1/√2 of its peak) is BW = f2 − f1 = f0/Q, which is R/(2πL) in series and 1/(2πRC) in parallel. The resonance frequency is the geometric mean f0 = √(f1f2), and f1,2 = f0(√(1 + 1/(4Q²)) ∓ 1/(2Q)). Losses also make the oscillation decay: in series RLC the amplitude falls as e−Rt/(2L).

Inductance from the bandwidth

L = R/(2π·BW) = 100/(2π·1000) = 0.01592 H

Capacitance from the resonance frequency

C = 1/((2π·10000)²·0.01592) = 1.592 × 10⁻⁸ F = 15.92 nF

Answer

C = 15.92 nF

Q = f₀/BW = 10.

9.Three-phase power

Answer: P = 6.4 kW

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Phase voltage

Vph = 400/√3 = 230.9 V

Phase current

|Z| = √(9² + 12²) = 15 Ω
Iph = 230.9/15 = 15.4 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·15.4²·9 = 6.4 kW

Answer

P = 6.4 kW

Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.

10.Three-phase power

Answer: QC = 38.16 kvar

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 120·tan(acos 0.78) = 120·0.8023 = 96.27 kvar
Q2 = 120·tan(acos 0.9) = 120·0.4843 = 58.12 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 38.16 kvar

Answer

QC = 38.16 kvar
11.RC & RL transients

Answer: i = 0.3162 A

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Time constant

τ = L/R = 220 mH / 33 Ω = 6.667 ms

Decay of the current

i(t) = 3·e−15/6.667 = 0.3162 A

Answer

i = 0.3162 A
12.RC & RL transients

Answer: vC = 0.2169 V

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Initial capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

vC(0) = 12·1/(4.7 + 1) = 2.105 V

Time constant after the switch opens

τ = R2·C = 1·220 = 220 ms

Discharge

vC(t) = 2.105·e−500/220 = 0.2169 V

Answer

vC = 0.2169 V
13.Thévenin & Norton

Answer: Rth = 6.4 Ω

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 3 ΩR3​ = 4 ΩSHORTR2​12 ΩAabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 3·12/(3+12) = 2.4 Ω
Rth = (R1 ∥ R2) + R3 = 2.4 + 4 = 6.4 Ω
14.Thévenin & Norton

Answer: IL = 1.538 A

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 25 ΩR3​ = 10 ΩSHORTR2​8 ΩAOPENBabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 25·8/(25+8) = 6.061 Ω
Rth = (R1 ∥ R2) + R3 = 6.061 + 10 = 16.06 Ω
R1​ = 25 ΩR3​ = 10 Ω+−V1​30 VR2​8 ΩAIs​3 ABab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 7.273 + 3·6.061 = 25.45 V
Vth = VB = VA + Is·R3 = 25.45 + 3·10 = 55.45 V
Rth​ = 16.06 Ω+−Vth​55.45 VRL​20 ΩIL​

Connect the load to the Thévenin equivalent

The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.

IL = Vth/(Rth + RL) = 55.45/(16.06 + 20) = 1.538 A
15.Bode plots & margins

Answer: ∠G(jω) = −100.2 °

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.

Add the angles

∠G = atan(2/20) − 90° − atan(2/7) = 5.711° − 90° − 15.95°

Result

∠G(j2) = −100.2°

Answer

∠G(jω) = −100.2°
16.Bode plots & margins

Answer: ωc = 14.32 rad/s

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.

Gain crossover

ω⁴ + 100ω² − 62500 = 0
ω² = (−100 + √(10000 + 4·62500))/2 = 205
ωc = 14.32 rad/s

Answer

ωc = 14.32 rad/s

ExamProblems · examproblems.com/courses/circuit-analysis-2